One millimole of Ni(NO_3)_2 dissolves in 280.0 mL of a solution that is 0.300 M

Question

One millimole of Ni(NO_3)_2 dissolves in 280.0 mL of a solution that is 0.300 M in ammonia. What is the initial concentration of Ni(NO_3)_2 in the solution? What is the equilibrium concentration of Ni^2+(aq) in the solution? The Ni^2+ forms a hexammonia complex: Ni(NH_3)_6^2+- K_f = 5.5 Times 10^8. Since complex ion formation constants are very large, it is almost a limiting reagent problem. It is easier to start by converting all of the Ni^2+ (the limiting reagent) into the ammonia complex and then letting a little of the complex (x M) turn back into Ni^2+ Very little will turn back: 6x

Explanation / Answer

Kf value for [Ni(NH3)6]2+ is 5.5×10^+8 – (from book)

Ni+2 & 6 NH3 --> [Ni(NH3)6]2+

Kf = [Ni(NH3)6]2+ / [Ni+2] [NH3]^6

5.5 x 10^8 = [0.00357] / [Ni+2] [0.300]^6

5.5 x 10^8 = [0.00357] / [Ni+2] (0.300)

[Ni+2] = [0.00357] / 5.5 e8 (0.0156)

Ni+2 = 4.16 x 10^-10 M

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